Proof The tangent spaces to the cylinders $\partial ^{+}C_{\varepsilon }$ and $\partial ^{-}C_{\varepsilon }$ at points $(x,y)$ are described via:

$$ \begin{align*} T_{(x,y)}\partial^{+}C_{\varepsilon}= \{(v_1,v_2)\in {\mathbb R}^n~|~\;\langle v_1,x\rangle=0\}, \end{align*} $$

$$ \begin{align*}T_{(x,y)}\partial^{-}C_{\varepsilon}= \{(v_1,v_2)\in {\mathbb R}^n~|~\;\langle v_2,y\rangle=0\}. \end{align*} $$

Hence, we need to look at

$$ \begin{align*}\langle X_1(x,y),x\rangle=\langle L^-x,x\rangle+\langle \tilde{f}(x,y)x,x\rangle\end{align*} $$

and

$$ \begin{align*}\langle X_2(x,y),y\rangle=\langle L^+y,y\rangle+\langle \tilde{g}(x,y)y,y\rangle.\end{align*} $$

We have that $\langle \tilde {f}(x,y)x,x\rangle = o(|x|^2)$ uniformly in y and $\langle \tilde {g}(x,y)y,y\rangle =o(|y|^2)$ uniformly in y since $\tilde {f}$ and $\tilde {g}$ are continuous and vanish at the origin.

The symmetry and definiteness of $L^-$ and $L^+$ imply now that there exists $\varepsilon _1$ and $\varepsilon _2$ such that

$$ \begin{align*} \langle L^-x,x\rangle+\langle \tilde{f}(x,y)x,x\rangle <0 \,\qquad \forall\;0<|x|\leq \varepsilon_1,\forall\; |y|\leq \varepsilon_1 \end{align*} $$

and

$$ \begin{align*}\langle L^+y,y\rangle+\langle \tilde{g}(x,y)y,y\rangle>0\,\qquad \forall\; 0<|y|\leq \varepsilon_2, \forall\; |x|\leq \varepsilon_2. \end{align*} $$

For $\varepsilon \leq \min \{\varepsilon _1,\varepsilon _2\}$ we get the transversality of X with both $\partial ^{\pm }C_{\varepsilon }$ .

Proof of (1). Let $q_1=(x_1,y_1)\prec q_2=(x_2,y_2), q_1,q_2\in {C}_{\varepsilon }$ . Let $\tau>0$ be the time needed to “travel” from $q_1$ to $q_2$ , i.e., $(x_2,y_2)=(x(\tau ,x_1,y_1),y(\tau ,x_1,y_1))$ . Now, consider the BVP defined by X with data $(x_1, y_2,\tau )$ . Since $|x_1,y_2|\leq \varepsilon $ , it follows from the Theorem 2.3 that for all $0\leq t\leq \tau $ , the following holds:

$$ \begin{align*} |x^{*}(t,x_1,y_2,\tau)| &\leq |x_1|\mathrm{e}^{-(\alpha-\delta)t}<|x_1|<\varepsilon,\\ |y^{*}(t,x_1,y_2,\tau)|&\leq |y_2|\mathrm{e}^{(\alpha-\delta)(t-\tau)}<|y_2|<\varepsilon. \end{align*} $$

Therefore, the portion of the trajectory comprised between $q_1$ and $ q_2$ is contained in ${C}_{\varepsilon }$ .

Proof of (2). Consider $(x_0,y_0)\in \partial ^{+}C_{\varepsilon }$ with $|y_0|<\varepsilon $ . The normal vectors v at $(x_0,y_0)$ to $\partial ^{+}C_{\varepsilon }$ that point to the interior of $C_{\varepsilon }$ are described by the inequality $\langle v_1,x_0\rangle <0$ and we already chose $\varepsilon $ so that the vector field X satisfies such an inequality. Hence, either the trajectory $(x(t,x_0,y_0),y(t,x_0,y_0))$ will belong to the interior of $C_{\varepsilon }$ for small $t>0$ or $(x_0,y_0)\in \partial ^+C_{\varepsilon }\cap \partial ^-C_{\varepsilon }$ i.e. $|y_0|=\varepsilon $ , which is not the case.

Suppose $t>0$ . For $y_0\neq 0$ we have $\displaystyle \lim _{t\rightarrow +\infty }(x(t),y(t))\neq (0,0)$ and the trajectory cuts again the boundary of $ C_{\varepsilon }$ . This happens because in the cube $C_{\varepsilon }$ the function $t\rightarrow |x(t)|$ is decreasing while $t\rightarrow |y(t)|$ is increasing. Indeed the derivatives of the square of these functions are equal to

$$ \begin{align*} \langle X_1(x,y),x\rangle \;\;\mbox{and}\;\; \langle X_2(x,y),y\rangle, \;\;\mbox{respectively} \end{align*} $$

and by our choice of $\varepsilon $ above the first one is negative while the second one is positive as long as they are not $0$ which would happen for a stable or unstable flow line.

Let $(x_1, y_1)\in \partial ^+C_{\varepsilon }\cup \partial ^-C_{\varepsilon }$ be the first point where this trajectory hits the boundary again and $ \tau>0$ the time needed to get from $(x_0,y_0)$ to $(x_1,y_1)$ .

Since $\langle X_1(x,y),x\rangle $ is a continuous function we see that $(x_1,y_1)\notin \partial ^{+}C_{\varepsilon }\setminus \mathrm {S}_{0}$ and moreover $|x_1|<\varepsilon $ , because

(2.10) $$ \begin{align}|x_1|=|x^{*}(\tau,x_0,y_1,\tau)| \leq |x_0|\mathrm{e}^{-(\alpha-\delta)\tau}<|x_0|\leq\varepsilon. \end{align} $$

Define $\mu $ , for the time being, as the map that associates to every $(x_0,y_0)\in \partial ^{+}C_{\varepsilon }\setminus (\mathrm {S}_{0}\cup \partial ^-C_{\varepsilon })$ the point of “first encounter” $(x_1,y_1)\in \partial ^{-}C_{\varepsilon }\setminus (\mathrm {U}_{0}\cup \partial ^+C_{\varepsilon })$ which lies on the same trajectory.

Suppose now that $|x_0|=|y_0|=\varepsilon $ . If there exist $\tau _0>0$ such that $(x(\tau _0,x_0,y_0),y(\tau _0,x_0,y_0))=(x_0^{\prime },y_0^{\prime })\in C_{\varepsilon }$ , define the BVP with data $(x_0,y_0^{\prime }, \tau _0)$ . Then, we get a contradiction from the estimates

$$ \begin{align*} \varepsilon= |y_0|=|y^*(0,x_0,y_0^{\prime},\tau_0)|\leq |y_0^{\prime}|\mathrm{e}^{-(\alpha-\delta)\tau_0}<|y_0^{\prime}|<\varepsilon. \end{align*} $$

Hence, for $|x_0|=|y_0|=\varepsilon $ , the trajectory determined by $(x_0,y_0)$ only intersects $C_{\varepsilon }$ in $(x_0,y_0)$ . Therefore, the natural extension of $\mu $ to $\partial ^+C_{\varepsilon }\cap \partial ^-C_{\varepsilon }$ is equal to the identity on this set.

The map $\mu $ is clearly bijective. The differentiability of $\mu $ is standard and proved along the following lines. The essential part is to prove the differentiability of the time-function that associates to each $(x_0,y_0)$ the time $t(x_0,y_0)$ it takes to get to $\mu (x_0,y_0)$ . Fix one such $(x_0,y_0)$ and use the diffeomorphism of the flow that corresponds to time $t(x_0,y_0)$ to flow a small open neighborhood of $(x_0,y_0)$ inside $\partial ^+C_{\varepsilon }$ which does not contain points in the stable manifold to an $n-1$ dimensional manifold H which contains $\mu (x_0,y_0)$ and is still transverse to X. Since $\mu (x_0,y_0)$ is not a critical point put coordinates in order to turn X into the generator of the first coordinate-translation by unit-time. One arrives thus at the problem of having to prove differentiability of the time-function obtained by going from one hypersurface to another hypersurface through the origin and both transverse to the first coordinate. That is simply the difference in height functions where the height is the first coordinate, hence a differentiable function.

Proof of (3). Suppose that there exists $0<\gamma \leq \varepsilon $ and a sequence $(x_{0n},y_{0n})$ such that $|x_{0n}|=\varepsilon $ , $|y_{0n}|\longrightarrow 0$ ( $y_{0n}\neq 0$ ) when $n\rightarrow \infty $ and $|x_{1n}|\geq \gamma $ for all n where $\mu (x_{0n},y_{0n})=(x_{1n},y_{1n})\in \partial ^-{C}_{\varepsilon }$ . Let $\tau _n>0$ be the sequence of moments such that

$$ \begin{align*}(x(\tau_n,x_{0n},y_{0n}),y(\tau_n,x_{0n},y_{0n}))=(x_{1n},y_{1n}).\end{align*} $$

It follows from the following estimates (see also (2.10))

(2.11) $$ \begin{align} \displaystyle\gamma\leq|x_{1n}|\leq |x_{0n}|\mathrm{e}^{-(\alpha-\delta)\tau_n}\leq\varepsilon\mathrm{e}^{-(\alpha-\delta)\tau_n} \end{align} $$

that $\displaystyle \tau _n\leq \ln \left (\frac {\varepsilon }{\gamma }\right )\frac {1}{\alpha -\delta }$ .

We can therefore select a convergent subsequence $(x_{0n_k}, \tau _{n_k})\longrightarrow (x_0^{\prime },\tau ^{\prime })$ . Note that $\tau ^{\prime }>0$ , because if $\tau ^{\prime }=0$ then $|y_{0n_k}|\longrightarrow \varepsilon $ . Indeed, the only points for which it takes $0$ -time to go from $\partial ^+C_{\varepsilon }$ to $\partial ^-C_{\varepsilon }$ are the points on $\partial ^+C_{\varepsilon }\cap \partial ^-C_{\varepsilon }$ and it is not hard to see that, due to the continuity of the time-function, the shorter the time it takes to get to $\partial ^-C_{\varepsilon }$ the closer to $\partial ^+C_{\varepsilon }\cap \partial ^-C_{\varepsilon }$ the starting point has to be.

Since $|y(\tau _{n_k},x_{0n_k},y_{0n_k})|=|y_{1n_k}|=\varepsilon $ we get that a contradiction with

$$ \begin{align*}y(\tau_{n_k},x_{0n_k},y_{0n_k})\rightarrow y(\tau^{\prime},x^{\prime},0)=0.\end{align*} $$

The later holds because the trajectory determined by $(x^{\prime },0)$ is stable.

Proof of (4). The flow-convex property is immediate from the analogous property of $C_{\epsilon }$ .

Suppose the $V^{\epsilon }_{\gamma }$ is not a neighborhood of $V_0$ . Hence, there exists a sequence $(x_n,y_n)\in C^{\epsilon }\setminus V_{\gamma }^\varepsilon $ with $(x_n,y_n)\longrightarrow (v_1,v_2)\in V_0^{\varepsilon }$ . Thus either $x_n\longrightarrow 0$ or $y_n\longrightarrow 0$ .

As none of the points $(x_n,y_n)$ is on $\mathrm {S}_0\cup \mathrm {U}_0$ there exist points of “first encounter” $(x_{0n},y_{0n})\in \partial ^+C_{\varepsilon }$ and $(x_{1n},y_{1n})\in \partial ^-C_{\varepsilon }$ such that

$$ \begin{align*}(x_{0n},y_{0n})\prec (x_n,y_n)\prec(x_{1n},y_{1n}). \end{align*} $$

Since $(x_n,y_n)\notin V^{\epsilon }_{\gamma }$ it follows that $|x_{1n}|\geq \gamma $ .

The case $x_n\rightarrow 0$ is disposed immediately by considering the BVP $(x_n,y_{1n},\tau _n)$ where $\tau _n$ is the time it takes to go from $(x_n,y_n)$ to $(x_{1n},y_{1n})$ . Then, we get a contradiction in

$$ \begin{align*}\gamma\leq|x_{1n}|\leq|x_{n}|e^{-(\alpha-\delta)\tau_n}\leq |x_{n}|\rightarrow 0.\end{align*} $$

For the case $y_n\rightarrow 0$ consider the BVP with data $(x_{0n},y_{1n},\tau _n^{\prime })$ where $\tau _n^{\prime }$ now is the time it takes to get from $(x_{0n},y_{0n})$ to $(x_{1n},y_{1n})$ . Recall that $|x_{0n}|=|y_{1n}|=\varepsilon $ . We obtain

$$ \begin{align*}\gamma\leq|x_{1n}|\leq|x_{0n}|e^{-(\alpha-\delta)\tau_n^{\prime}} \; \Rightarrow \; \tau_n\leq \ln{\left(\frac{\varepsilon}{\delta}\right)}\frac{1}{\alpha-\delta}.\end{align*} $$

$$ \begin{align*}|y_{0n}|\leq |y_n|\mathrm{e}^{-(\alpha-\delta)s_n}\leq |y_n|,\end{align*} $$

where $0<s_n\leq \tau _n^{\prime }$ . Since $|y_n|\longrightarrow 0$ , we have $y_{0n}\longrightarrow 0$ .

Now, since $|x_{0n}|=\varepsilon $ and $\tau _n^{\prime }$ are bounded and $y_{0n}\rightarrow 0$ , we are in the same scenario that led to a contradiction in the proof of item (3).

In order to prove that $\lim _{\gamma \rightarrow 0} V^{\varepsilon }_{\gamma }=V_0^{\varepsilon }$ as defined in the statement it is enough to prove that

(2.12) $$ \begin{align} \bigcap V^{\varepsilon}_{\gamma_n}=V_0^{\varepsilon} \end{align} $$

for any decreasing sequence $\gamma _{n}\rightarrow 0$ . Indeed if (2.12) holds then fix one such sequence $\gamma _n$ . Suppose $\exists u_n\in V^{\varepsilon }_{\gamma _n}\setminus U$ for all n. Since $\overline {V^{\varepsilon }_{\gamma _{n+1}}}\subset V^{\varepsilon }_{\gamma _n}$ and $\overline {V^{\varepsilon }_{\gamma _{n+1}}}$ (the closure in $C_{\varepsilon }$ ) is compact we get that we can extract a convergent subsequence $u_{n_k}$ whose limit necessarily belongs to $\bigcap V^{\varepsilon }_{\gamma _n}=V_0^{\varepsilon }$ . But this is a contradiction with U being a neighborhood of $V_0^{\varepsilon }$ .

In order to prove (2.12), take a point $a=(x,y)\in \bigcap V^{\varepsilon }_{\gamma _n}$ such that $x\neq 0\neq y$ . Since $y\neq 0$ it follows that $a\notin S_0$ and therefore a lies on a trajectory which hits $\partial ^-C_{\varepsilon }$ in a point $b=(x^{\prime },y^{\prime })\notin U_0$ , i.e., $x^{\prime }\neq 0$ . But $a\in \bigcap V^{\varepsilon }_{\gamma _n}$ implies that $|x^{\prime }|<\gamma _n$ for all n, hence $x^{\prime }=0$ . Contradiction.▪

Proof (Sketch) The idea is to turn to BVP coordinates via (2.4) in

$$ \begin{align*} W=\left\{\left(\tau, x_1^*\left(x_0,y_1,\frac{\tau}{1-\tau}\right),y_1,x_0,y_0^*\left(x_0,y_1,\frac{\tau}{1-\tau}\right)\right)~\bigg|~|x_0,y_1|<\varepsilon, \tau\in(0,1)\right\}. \end{align*} $$

Use now (2.9) to conclude that $x_1^*$ and $y_0^*$ converge uniformly to $0$ when $\tau \rightarrow 1$ together with their derivatives, obtaining thus that W is the graph of a smooth function over $\mathring {C}_{\varepsilon }\times [0,1]$ .▪

Proof In the cube $C_{\varepsilon }$ of Theorem 2.4, there exists $0<\gamma <\varepsilon $ such that

$$ \begin{align*}\inf_{u\in T_{\gamma}}{f}(u)>0,\end{align*} $$

where $T_{\gamma }=\{(x,y)\in \partial ^{-}{C}_{\varepsilon }; |x|<\gamma , |y|=\varepsilon \}$ since $f>0$ on the compact $\partial ^{-}{C}_{\varepsilon }\cap U_{{\mathbf {p}_{\mathbf {1}}}}$ . Fix such a $\gamma $ .

Choose now $\theta>0$ with

(4.1) $$ \begin{align} 0<\theta< \inf_{u\in T_{\gamma}}f(u). \end{align} $$

Notice that each trajectory that starts inside $V^{\varepsilon }_{\gamma }$ either ends up at the critical point or leaves $V^{\varepsilon }_{\gamma }$ through $T_{\gamma }$ . Let $p\in V^{\varepsilon }_{\gamma }\cap f^{-1}(-\infty ,\theta ]$ . On one hand, f is increasing and continuous along the trajectories and on the other hand, $V^{\epsilon }_{\gamma }\cap \gamma _p$ Footnote ¹ is connected by the flow-convex property of $V^{\varepsilon }_{\gamma }$ . It follows that $\gamma _p$ meets $f^{-1}(\theta )$ before reaching $T_{\gamma }$ by (4.1).

Part (b) is analogous.

Since $s_0(b_0)\in S_{{\mathbf {p}_{\mathbf {1}}}}$ we have that $f(s(b_0))\leq 0$ and therefore one can choose $B_0$ with $s_0(B_0)\subset V_{\gamma }^{\epsilon }\cap f^{-1}(-\infty ,\theta ^{\prime })$ .

Part (ii) is an immediate consequence of (i) and the first part of the proof.▪

Proof For any regular value $\theta $ the intersection $U\cap {f}^{-1}(-\infty , \theta ]$ is a manifold with boundary for any open $U\subset M$ such that $f^{-1}(\theta )\cap U\neq \emptyset $ .▪

Proof Notice that $\mathring {\overline {W}_1}=W_1$ .

We start with the intersection ${W}_1\cap \mathring {Z}_{\theta }$ . The transversality is immediate from the fact that $W_1$ is a graph over the first plus the last two variables, i.e., ${\mathbb R}\times \mathring {{{C}_{\varepsilon }}}\times B_0$ while the first component of $\mathring {Z}_{\theta }$ is an open subset of $\mathring {{{C}_{\varepsilon }}}$ .

A similar reasoning applies for $q\in \partial _0 \overline {W_1}\cap \mathring {Z}_{\theta }$ . In fact, due to the transversality of the flow to a regular level, set $f^{-1}(\theta )$ this also proves the transversality of ${W_1}$ with $\partial {Z}_{\theta }$ .

Notice that $\partial _0{\overline {W_1}}\cap \partial {Z}_{\theta }=\emptyset $ since a point $q\in \xi _0(B_0)$ cannot satisfy $\tilde {f}(q)=\theta $ due to property (i) of Proposition 4.1.

The transversality of $\partial _1\overline {W_1}$ with $\mathring {Z_1}$ follows from the transversality of $s(B_0)$ and $S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0$ .

Finally, the transversality of $\partial _1\overline {W_1}$ with $\partial Z_1$ follows from the transversality of $\mathrm {U}_{{\mathbf {p}_{\mathbf {1}}}}$ and $\tilde {f}^{-1}(\theta )$ together with the transversality of $s(B_0)$ and $S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0$ .▪

Proof We use a standard trick. Clearly, $N_1\times N_2$ is a manifold with corners of type $k+l$ . Then the complete transversality of $N_1$ and $N_2$ is equivalent with the complete transversality of $N_1\times N_2$ and the diagonal submanifold $\Delta $ in $N\times N$ . The conclusion then is that $(N_1\times N_2)\cap \Delta $ is a manifold with corners (for this see Proposition A.3 in [Reference Cibotaru7]).▪

Proof Recall that for locally compact metric spaces $X,Y$ , a map $F:X\rightarrow Y$ is proper if and only if for any sequence $(x_n)_{n\in {{\mathbb N}}}\in X$ such that $\displaystyle \lim _{n\rightarrow \infty }x_n=\infty $ one has $\displaystyle \lim _{n\rightarrow \infty } F(x_n)=\infty $ . By definition,

$$ \begin{align*} \lim_{n\rightarrow \infty} x_n=\infty \end{align*} $$

if for every $K\subset X$ compact there exists $n_0\in {{\mathbb N}}$ such that $x_n\in X\setminus K$ for all $n\geq n_0$ . Notice that such a sequence does not have any convergent subsequence in X and the converse is also true. It follows easily then that a map $F:X\rightarrow Y$ is not proper if and only if there exists $(x_n)_{n\in {{\mathbb N}}}\in X, x_n\rightarrow \infty $ such that $F(x_n)\rightarrow y\in Y$ .

Assume therefore that $(u_n)_{n\in {{\mathbb N}}}\in \mathcal {A}_1$ satisfies $u_{n}\rightarrow \infty $ while $\mathcal {R}(u_{n})\rightarrow \tilde {u}$ . Now $u_n$ has $4$ components:

(4.6) $$ \begin{align} u_n=(t_n,a_n^{\prime},a_n,b_n)\in [0,1]\times V_{\theta}\times V_{\theta}\times B_0. \end{align} $$

By passing to a subsequence of $u_n$ , we can assume $t_n$ converges. There are two possibilities.

Either $t_n\rightarrow t^{\prime }<1$ or $t_n\rightarrow 1$ . We analyze them separately.

First, the triple $\mathcal {R}(u_{n})=(a_n^{\prime },a_n,b_n)$ converges in $V_{\theta }\times V_{\theta }\times B_0$ . From $u_n\in Z_1$ we get $(a_n,b_n)=(s_0(\beta _n),\beta _n)$ for some $\beta _n\in B_0$ . Hence, $\beta _n$ converges to $\beta \in B_0$ and $s_0(\beta _n)\rightarrow s_0(\beta )$ .

If $t_n\rightarrow t^{\prime }<1$ then for n big enough we have that $u_n\in \overline {W_1}\setminus \{t=1\}$ and therefore

$$ \begin{align*}a_n^{\prime}=\psi_{\frac{t_n}{1-t_n}}(a_n). \end{align*} $$

Hence $a_n=s_0(\beta _n)\rightarrow s_0(\beta )$ and $a_n^{\prime }\rightarrow \psi _{\frac {t^{\prime }}{1-t^{\prime }}}(s_0(\beta ))$ and by hypothesis this is in $V_{\theta }$ . We conclude that

$$ \begin{align*} u_n\rightarrow \left(t^{\prime}, \psi_{\frac{t^{\prime}}{1-t^{\prime}}}(s_0(\beta)),s_0(\beta),\beta\right) \end{align*} $$

and this limit belongs to $Z_1\cap \overline {W_1}=\mathcal {A}_1$ . Contradiction with $u_n\rightarrow \infty $ .

If $t_n\rightarrow 1,$ we have that $u_n=(t_n,a_n^{\prime },s_0(\beta _n),\beta _n)$ and the convergence of $(a_n^{\prime },s_0(\beta _n),\beta _n)$ to a point in $V_{\theta }\times V_{\theta }\times B_0$ implies again that $\beta _n\rightarrow \beta \in B_0$ . Since $s_0(B_0)\subset V_{\theta }$ , we have that $s_0(\beta _n)\rightarrow s_0(\beta )\in V_{\theta }$ . We get therefore that the sequence $u_n$ converges in $[0,1]\times V_{\theta }\times V_{\theta }\times B_0 $ since all its coordinates converge. In order to reach a contradiction, we need only check that it converges to some element u of $Z_1\cap \overline {W_1}$ . We have that

$$ \begin{align*}u_n\rightarrow u\in {\mathbb R}\times\mathring{{{C}_{\varepsilon}}}\times \mathring{{{C}_{\varepsilon}}}\times B_0.\end{align*} $$

On the other hand, since $u_n\in \overline {W_1}$ and the closure of $W_1$ is taken within $ {\mathbb R}\times \mathring {{{C}_{\varepsilon }}}\times \mathring {{{C}_{\varepsilon }}}\times B_0$ we get that $u\in \overline {W_1}$ . One sees easily that $u\in Z_1$ since $(a_n^{\prime },s_0(\beta _n),\beta _n)=\mathcal {R}(u_n)$ converges to a point $(a^{\prime },s_0(\beta ),\beta )\in V_{\theta }\times V_{\theta }\times B_0$ .▪

Proof At (1), let $q\in \overline {W_1}\cap Z_1$ . On one hand

$$ \begin{align*} q=\lim_{n\rightarrow \infty}\left (t_n,\psi_{\frac{t_n}{1-t_n}}\left(x_n,y_n\right),x_n,y_n,b_n\right), \end{align*} $$

where $b_n\in B$ , $(x_n,y_n)\in \mathring {C}_{\varepsilon }$ , $t_n\in [0,1]$ . From $q\in Z_1$ it follows that $(x_n,y_n,b_n)\rightarrow (s_0(\beta ),\beta )$ . Suppose $t_n\rightarrow t^{\prime }<1$ . Then

$$ \begin{align*}\psi_{\frac{t_n}{1-t_n}}(x_n,y_n)\rightarrow \psi_{\frac{t^{\prime}}{1-t}}(s_0(\beta)). \end{align*} $$

Hence, in this case, $\mathcal {R}(q)=\left (\psi _{\frac {t^{\prime }}{1-t^{\prime }}}(s_0(\beta )),s_0(\beta ),\beta \right )\in \xi _{\frac {t^{\prime }}{1-t^{\prime }}}(B_0)$ .

When $t_n\rightarrow 1$ , $q\in \overline {W_1}$ implies that $(x_n,y_n,b_n)\rightarrow (s_0(\beta ),\beta )\in S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0$ , i.e., $\beta \in s^{-1}(S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0)$ and $\psi _{\frac {t_n}{1-t_n}}\left (x_n,y_n\right )\rightarrow q_1\in U_{{\mathbf {p}_{\mathbf {1}}}}\cap V_{\theta }$ .

We argue that due to the transversality of s with $S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0$ all points in $U_{{\mathbf {p}_{\mathbf {1}}}}\cap (V_{\theta }\times s(s^{-1}(S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0)))$ are limits of type $\xi _{t_n}(s_0(\beta _n),s_0(\beta _n),\beta _n)$ , $t_n\geq 0$ . Fix first $q_2=(x_2,0,b_2)\in s(s^{-1}(S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0))$ . Since $s\pitchfork B_0$ we can provide a submanifold $B_0^{\prime }\subset B_0$ of dimension equal to $\dim {U}_{{\mathbf {p}_{\mathbf {1}}}}$ such that $s_0\big |_{B_0^{\prime }}\pitchfork S_{{\mathbf {p}_{\mathbf {1}}}}$ and $q_2\in s(B_0^{\prime })$ . Take then a transverse small disk D in $q_2+T_{q_2}s(B_0^{\prime })$ centered at $q_2$ . The trajectories originating in this disk will cut $s(B_0^{\prime })$ exactly once due to the transversality of $s(B_0^{\prime })$ to the flow. This stays true even if $q_2=(0,0,b_2)$ is critical. By Corollary 2.9 which can be applied also to “slanted” disks, given any point $(0,y_2)\in U_{{\mathbf {p}_{\mathbf {1}}}}$ there exists a sequence of points $u_n\in D$ with $u_n\rightarrow q_2$ and a corresponding sequence of points on trajectories determined by $u_n$ that converges to $(0,y_2)$ .

This finishes the inclusion $\mathcal {R}(\mathcal {A}_1)\subset \overline {\bigcup _{t\geq 0}\xi _t(B_0)}$ .

Conversely, let $ (a,b,c)\in \overline {\bigcup _{t\geq 0}(\xi _t(B_0))}\bigcap \tilde {V}_{\theta }$ . Then there exist $t_n\geq 0$ and $b_n\in B_0$ such that

$$ \begin{align*}\lim_{n\rightarrow\infty}(\psi_{t_n}(s_1(b_n)),s_1(b_n), b_n)= (a,b,c)\end{align*} $$

with $c\in B_0$ . By passing to a subsequence one can assume that $t_n\rightarrow t_0$ or $t_n\rightarrow \infty $ . Since $a\in V_{\theta }\subset \mathring {V}^{\varepsilon }_{\gamma }$ and the latter is open in M we can consider $\psi _{t_n}(s_1(b_n))\in \mathring {V}^{\varepsilon }_{\gamma }$ for n big enough. However, $\psi _{t_n}(s_1(b_n))$ might not be in $V_{\theta }$ for infinitely many n, since it could happen that $f(\psi _{t_n}(s_1(b_n)))>\theta $ for a subsequence.

Let $r_n:=\frac {t_n}{1+t_n}$ . We have that

$$ \begin{align*} u_n:=\left(r_n,\psi_{t_n}(s_1(b_n)),s_1(b_n), b_n\right)\in W_1. \end{align*} $$

Since $r_n$ , $b_n$ and $\psi _{t_n}(s_1(b_n))$ all converge we have that in fact $u_n$ converges to a point $u\in [0,1]\times {V}_{\theta }\times \mathring {{C_{\varepsilon }}}$ that necessarily lies in $\overline {W_1}$ . The limit $u=(r,a,b,c)$ will also be a point in $Z_1$ since $(b,c)\in s(B_0)$ and $a\in V_{\theta }$ . Hence, $(a,b,c)\in \mathcal {R}(\mathcal {A}_1)$ .

Statements (2) and (3) are trivial from the description of $\partial ^{1}_{1}\mathcal {A}_1$ and $\partial _0^1\mathcal {A}_1$ .

For (4) the inclusion $\subset $ is straightforward. For the inclusion $\supset $ notice that if $(a,b,c)\in \mathcal {A}_1$ then either ${f}(a)<\theta $ in which case $a\in \mathring {{V_{\theta }}}$ and so $(a,b,c)\in \mathring {Z_1}\cap \overline {W_1}$ or ${f}(a)=\theta $ in which case $(a,b,c)\in \partial {Z_1}\cap \overline {W_1}=:\partial ^1_2\mathcal {A}$ .

For (5) one notices that for $t\neq 1$ the (restriction of the) map $\mathcal {R}$ is injective away from the points corresponding to $s^{-1}(\{{\mathbf {p}_{\mathbf {1}}}\}\times B_0)$ . Moreover, $\mathcal {R}$ is injective when $t=1$ . For $p\neq q$ with $t_p\neq 1$ and $t_q=1$ , $\mathcal {R}(p)\neq \mathcal {R}(q)$ unless $p,q\in \overline {W}_{{\mathbf {p}_{\mathbf {1}}}}$ .

For (6) one notices that $\overline {W}_{{\mathbf {p}_{\mathbf {1}}}}\cap \partial ^1_2\mathcal {A}_1=\emptyset $ .▪

Proof The intersection $\mathring {\mathcal {A}}_{1}:=\mathcal {A}_1\cap (\mathring {V}_{\theta }\times M\times B_0)$ is a manifold with boundary since $\partial ^1_2\mathcal {A}_1$ gets eliminated in the intersection. Since $\mathcal {R}$ is proper we can push-forward any current. Use:

$$ \begin{align*} d (\mathcal{R}_*(\mathring{\mathcal{A}}_{1}))=\mathcal{R}_*(d\mathring{\mathcal{A}}_{1})=\mathcal{R}_*(\partial^1_1\mathcal{A}_1)-\mathcal{R}_*(\partial^1_0\mathcal{A}_1). \end{align*} $$

Now $\mathcal {R}\big |_{\mathring {\mathcal {A}}_{1}}$ away from a set of zero measureFootnote ² is a bijection onto

$$ \begin{align*}\xi([0,\infty)\times B_0)\cap \mathring{V}_{\theta}\times M\times B_0.\end{align*} $$

It follows from the area formula that

$$ \begin{align*}\mathcal{R}_*(\mathring{\mathcal{A}}_{1})=T\big|_{\mathring{V}_{\theta}\times M\times B_0}, \end{align*} $$

where T is the current appearing in (3.4).▪

Proof Since $\partial _{2}^{1}\mathcal {A}_1$ is a manifold with boundary $\partial ^2\mathcal {A}_1$ , as defined in (4.5), we need to show transversality at points

1. (1) $q\in \sigma (\partial ^2\mathcal {A}_1)\cap (\mathrm {S}_{{\textbf {p}}_2}\times M\times B_0) $ and

2. (2) $q\in \sigma (\partial _{2}^{1}\mathcal {A}_1\setminus \partial ^2\mathcal {A}_1) \cap (\mathrm {S}_{{\textbf {p}}_2}\times M\times B_0).$

For both situations, take the unique (by (6) of Proposition 4.6) $q^{\prime }\in \partial _{2}^{1}\mathcal {A}_1$ such that $\sigma (q^{\prime })=q$ . The two situations are distinguished by $t_{q^{\prime }}=1$ (for (1)) or $t_{q^{\prime }}\neq 1$ .

From the explicit expression of $\partial ^2\mathcal {A}_1$ in (4.5), we see that transversality for (1) is implied by the Smale property of the flow.

For (2) since $t\neq 1$ , we can give another description to $\overline {W_1}\setminus \{t=1\}\cap (f^{-1}(\theta )\times M\times B_0)$ as the graph of the time map defined over $\xi _0(B_0\setminus s^{-1}(S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0))$ that associates to a point p the time $t_p$ it needs to reach $\tilde {f}^{-1}(\theta )$ . We deduce that the intersection of this time map graph with the flow-invariant $S_{p_2}\times M\times B_0$ can be described as the flow-out of the intersection $\xi _0(B_0\setminus s^{-1}(S_{{\mathbf {p}_{\mathbf {1}}}}\times B_0))\cap (S_{{\textbf {p}}_2}\times M\times B_0)$ to the level set $\tilde {f}^{-1}(\theta )$ . By the transversality of s with $S_{p_2}$ , we get that this intersection is transverse. Moreover, since the flow preserves transversality, we get by Proposition 2.10 the transversality condition we are after within $\tilde {f}^{-1}(\theta )$ .▪

Proof Analogous to the proof of Lemma 4.3.

The reason for (4.14) is that the f-value of the second component of $\overline {W_2}$ is at least as big as the f-value of the third component while this is not the case for an element of $\partial ^1_0Z_{\sigma }$ due to $f \circ \sigma _1=-\theta ^{\prime }>-\theta _1$ .▪

Proof Suppose, just like in Proposition 4.5 that there exists a sequence $u_n\in \mathcal {A}_{\sigma } $ such that $u_n\rightarrow \infty $ and $\mathcal {R}_{\sigma }(u_n)$ converges. Now, $u_n$ has six components

$$ \begin{align*} u_n=(t_n,a_n^{\prime},a_n,m_n,b_n, z_n) \end{align*} $$

with $t_n\in [0,1]$ . By passing to a subsequence we can assume that $t_n\rightarrow t^{\prime }\in [0,1]$ . We will show that in both cases $t^{\prime }=1$ and $t^{\prime }\neq 1$ one reaches a contradiction by showing that there exists a subsequence of $u_n$ which converges in $\mathcal {A}_{\sigma }$ .

For the case $t^{\prime }=1$ , we will use the fact that $\{t=1\}\cap \overline {W}\cap ([0,1]\times V_{\theta _1}\times V_{\theta _1})$ is compact and thus has a compact neighborhood. In fact, one notices that $\overline {W}$ is completely transverse to $V_{\theta _1}\times V_{\theta _1}\times {\mathbb R}$ and their intersection is a manifold with corners, with one of the codimension $1$ boundaries being contained in $\{t=1\}$

$$ \begin{align*} \bigcup_{{\textbf{p}}\in\operatorname{\mathrm{Crit}}(f)\cap f^{-1}(0)}\{1\}\times(U_{\textbf{p}}\cap V_{\theta_1})\times(S_{\textbf{p}}\cap V_{\theta_1}). \end{align*} $$

This is compact. We conclude that from $(a_n^{\prime },a_n)$ we can extract a subsequence denoted again $(a_n^{\prime },a_n)$ that converges to a point in $V_{\theta _1}\times V_{\theta _1}$ . On the other hand, we have by hypothesis that $(a_n^{\prime },m_n,b_n)$ converges. Hence by passing to a subsequence we conclude that $(a_n,m_n,b_n)$ converges to some point $(a,m,b)$ . But $(a_n,m_n,b_n)=(\sigma _1(z_n),\sigma _2(z_n),\sigma _3(z_n))$ and $\sigma $ is proper. It follows that we can extract yet another subsequence this time from $z_n$ that converges to $z\in N$ (just take $\sigma ^{-1}(K)$ where K is a compact neighborhood of $(a,m,b)$ ). But then by the continuity of $\sigma $ we get that $(a_n,m_n,b_n, z_n)$ converges to $(a,m,b,z)\in \Gamma _{\sigma }$ . Since $a_n^{\prime }$ converges to a point in $V_{\theta _1}$ we conclude that $u_n$ converges to a point in $\mathcal {A}_{\sigma }$ , contradiction with $u_n\rightarrow \infty $ .

For $t^{\prime }\neq 1$ , we use that $a_n^{\prime }=\psi _{\frac {t_n}{1-t_n}}(a_n)$ and since $a_n^{\prime }$ converges and $\frac {t_n}{1-t_n}\rightarrow \frac {t^{\prime }}{1-t^{\prime }}$ , we conclude that $a_n$ also converges. We claim that $a_n$ converges to $a\in V_{\theta _1}$ . First $a_n^{\prime }\rightarrow a^{\prime }\in V_{\theta _1}$ by hypothesis. Now $V_{\theta _1}$ is a flow-convex neighborhood and $a^{\prime }=\psi _{\frac {t^{\prime }}{1-t^{\prime }}}(a)$ . It follows that $a\in V_{\theta _1}$ . Now the contradiction is obtained as before $(a_n,m_n,b_n)=(\sigma _1(z_n),\sigma _2(z_n),\sigma _3(z_n))$ , and so on.▪

Proof We need only be concerned with items (ii)–(iv). The distinguished boundary is defined as:

$$ \begin{align*} N^{\prime}:=\partial_{\theta_1}^1\mathcal{A}_{\sigma}:=\overline{W_2}\cap \partial^1_1Z_{\sigma}=\overline{W_2}\cap( [0,1]\times {f}^{-1}(\theta_1)\times \Gamma_{\sigma}). \end{align*} $$

Clearly $\mathcal {R}_{\sigma }({\partial _{\theta _1}^1\mathcal {A}_{\sigma }})\subset \tilde {f}^{-1}(\theta _1)$ .

In order to prove (ii), it is useful to consider the projection $\tilde {\mathcal {R}}_{\sigma }$ from $\mathcal {A}_{\sigma }$ onto the second, fourth, fifth, and sixth components. Then the image of this map will give the closure of the flow-out of $\Gamma _{\sigma }$ inside $V_{\theta _1}\times M\times B_0\times {\mathbb R}^j$ and the proof of this fact follows the same lines as the proof of item (1) in Proposition 4.6. Then projecting the closure of the flow-out of $\Gamma _{\sigma }$ onto $V_{\theta _1}\times M\times B_0$ equals $\overline {\left (\bigcup _{t\geq 0}\xi _t(\sigma (N))\right )}\cap \tilde {V}_{\theta _1}$ and this takes care of (ii).

The map $\tilde {R}_{\sigma }$ mentioned in the previous paragraph is injective on $\mathcal {A}_{\sigma }\setminus \{t=1\}$ since on this set all points are of type

$$ \begin{align*} (t,\psi_{\frac{t}{1-t}}(\sigma_1(z)),\sigma_1(z),\sigma_2(z),\sigma_3(z),z),\qquad z\in N, t\in [0,1), \end{align*} $$

which get projected to $(\psi _{\frac {t}{1-t}}(\sigma _1(z)),\sigma _2(z),\sigma _3(z),z)$ . Clearly all the points in the forward-flowout of $\Gamma _{\sigma }$ that are inside $V_{\theta _1}\times M\times B_0\times {\mathbb R}^j$ are also in the image of this projection. In order to obtain a set where $\mathcal {R}_{\sigma }$ is injective one needs to take out more points.

Since $\sigma \big |_{\partial ^2N}$ is completely transverse when restricted to all distinguished boundary pieces $\partial ^2_i N$ of $\partial ^2N$ , it follows that, we can define a natural subspace $\mathcal {A}_{\sigma \big |_{\partial ^2N}}$ of $\mathcal {A}_{\sigma }$ by taking the union of the corresponding sets $ \overline {W_2}\cap Z_{\sigma \big |_{\partial ^2_iN}}$ . Since this $\mathcal {A}_{\sigma \big |_{\partial ^2N}}$ will be a union of manifolds with corners of lower dimension it will have measure zero inside $\mathcal {A}_\sigma $ . Then $\mathcal {R}_{\sigma }$ will be injective on $\mathcal {A}_{\sigma }\setminus \left (\{t=1\}\cup \mathcal {A}_{\sigma \big |_{\partial ^2N}}\right )$ and this takes care of (iii).

In order to prove transversality, we can use Lemma 4.17 in order to reduce to the proof of transversality of $\tilde {\mathcal {R}}_{\sigma }\big |_{\partial ^1_{\theta _1}\mathcal {A}_{\sigma }}$ with $S_p\times M\times B_0\times {\mathbb R}^j$ . This then follows the same scheme as Lemma 4.8.

For injectivity, we note that $ \partial ^1_0\overline {W_2}\cap \partial ^1_1Z_{\sigma }=\emptyset $ and we separate $ N^{\prime }=\partial ^1_{\theta _1}\mathcal {A}_{\sigma }$ into two parts:

$$ \begin{align*} N_1:=(\overline{W_2}\setminus \{t=1\})\cap \partial^1_1Z_{\sigma} \;\;\mbox{and}\;\; N_2:=\partial^1_1\overline{W_2}\cap\partial^1_1Z_{\sigma}. \end{align*} $$

We observe that $\mathcal {R}_{\sigma }$ takes $N_1$ and $N_2$ to two disjoint sets in $\tilde {V}_{\theta }$ distinguished by the fact that the first component belongs to $\cup _{p\in \operatorname {\mathrm {Crit}}(f)\cap f^{-1}(0)} U_p$ (in the case of points in $\mathcal {R}_{\sigma }(N_2)$ ) or does not belong to the same set (for $\mathcal {R}_{\sigma }(N_1)$ ). We have that $N_2$ is a distinguished boundary of $N^{\prime }$ while $N_1$ contains the top stratum of $N^{\prime }$ . Hence it is enough to prove the injectivity of $\mathcal {R}_{\sigma }$ separately on $N_1\setminus \partial ^2 N_1$ and on $N_2\setminus \partial ^1N_2$ .

The points in $N_1\setminus \partial ^2 N_1$ are of type $\left (t_{0},\psi _{\frac {t_0}{1-t_0}}(\sigma _1(z)),\sigma _1(z),\sigma _2(z),\sigma _3(z),z\right )$ with $z\in N\setminus \partial ^2N$ and they get mapped to $(a,m,b):=\left (\psi _{\frac {t_0}{1-t_0}}(\sigma _1(z)),\sigma _2(z),\sigma _3(z)\right )\in f ^{-1}(\theta _1)\times M\times B_0$ . Then $t_0$ is the unique time it takes to flow backwards from point $a\in f ^{-1}(\theta _1)$ to level $f^{-1}(-\theta ^{\prime })$ , $\sigma _1(z)$ is the point of intersection with level $f^{-1}(-\theta ^{\prime })$ of the trajectory determined by a, and $z\in N\setminus \partial ^2N$ is uniquely determined by $\sigma (z)$ due to the hypothesis.

The description $N_2=\displaystyle \bigcup _{{\textbf {p}}\in f^{-1}(0)\cap \operatorname {\mathrm {Crit}}(f)}\{1\}\times (U_{{\textbf {p}}}\cap f^{-1}(\theta _1))\times \Gamma _{\sigma \big |_{\sigma ^{-1}(S_{\textbf {p}}\times M\times B_0)}}$ is useful. For the top stratum $(N_2)^0$ of $N_2$ , one restricts $\Gamma _{\sigma }$ to $\sigma ^{-1}(S_{\textbf {p}}\times M\times B_0)\cap N^0$ . If $p=(u,\sigma _1(z),\sigma _2(z),\sigma _3(z),z)\in (N_2)^0$ then u does not determine $\sigma _1(z)$ anymore, but property (d) says that $\sigma _1(z)$ is determined by $(\sigma _2(z),\sigma _3(z))$ . Since $\sigma $ is also injective on $N^0$ we get that p determines z. Hence, the map on $(N_2)^0$ that takes p to $(u,\sigma _2(z),\sigma _3(z))$ is injective and this finishes the proof of this issue.

Finally, property (d) itself holds for $\sigma ^{\prime }:N^{\prime }\rightarrow \tilde {V}_{\theta _1}\cap ({f}^{-1}({\theta _1})\times M\times B_0)$ . This follows from the description of $(N^{\prime })^0=(N_1)^0$ above and the fact that $\sigma $ is injective on $N^0$ and also satisfies property (d).▪

Proof Straightforward.▪

Proof One uses Lemma 4.14, Proposition 4.9 and Stokes on the manifold with corners $\mathcal {A}_{\sigma }$ pushed-forward via the proper map $\mathcal {R}_{\sigma }$ . The injectivity property identifies $(\mathcal {R}_{\sigma })_*(\mathcal {A}_{\sigma })$ with $T_{\sigma }$ and $(\mathcal {R}_{\sigma })_* (\overline {W_2}\cap \partial ^1_{j+1}Z_{\sigma })$ with $T_{\sigma \big |_{\partial ^1_j N}}$ for all j.▪

Proof of Theorem 3.2

We have already sketched the proof strategy at the end of Section 3. We need to explain what are the open sets $M_j\times M\times B_0$ that cover $M\times M\times B_0$ where the currential identity is true. We will discuss only the situations where $B_0$ is a small neighborhood around a point $b_0$ such that the forward trajectory determined by $s(b_0)$ ends at a nonmaximal point. The remaining situation was already discussed in Remark 3.6.

Let $c_0<\cdots <c_l$ be the consecutive critical levels of f excluding the maximum with $c_0$ the first encountered critical level by the forward trajectory of $s(b_0)$ . It might even be a minimal level of f, i.e., a level that contains a local minimum if $s(b_0)$ is a local minimum. Let $c_0<\delta _1<\delta _2<c_1<\delta _3<\delta _4<\cdots <c_{l-1}<\delta _{2l-1}<\delta _{2l}<c_l< \delta _{2l+1}$ be regular level sets such that $\delta _1=c_0+\theta _0$ and for $k\geq 1$ , $\delta _{2k}=c_{k}-\theta _k$ and $\delta _{2k+1}=c_k+\theta _k$ where $\theta _k$ is chosen small enough so that we can chose neighborhood $V_{\theta _k}$ around the critical points of level $c_k$ satisfying the conditions of Proposition 4.1.

Corollary 4.7 takes care of the first step of induction and implies the formula for

$$ \begin{align*}{f}^{-1}(-\infty,\delta_1)\times M\times B_0.\end{align*} $$

By property (4) of Proposition 4.6 and Lemma 4.14 there exists a map for some $\epsilon _1>0$ :

$$ \begin{align*}\sigma:\partial^1_2\mathcal{A}_1\rightarrow {f}^{-1}(\delta_1-\epsilon_1)\times M\times B_0,\end{align*} $$

whose image contains the closure of the forward-flow of $\xi _0(B_0)$ intersected with level $\delta _1-\epsilon _1$ and is transverse to all $S_{\textbf {{p}}}\times M\times B_0$ for all critical ${\textbf {p}}$ . Moreover, $\partial ^1_2\mathcal {A}_1$ is a manifold with boundary $\partial ^2\mathcal {A}_1=U_{{\textbf {p}}_1}\cap {f}^{-1}(\delta _1-\epsilon _1)\times s(s^{-1}(S_{{\textbf {p}}_1})\times B_0)$ and $\sigma $ satisfies the conditions of Proposition 4.9.

Let $T^1_{\sigma }$ to be the current determined by the flow-out of the image of $\sigma $ in the open set ${f}^{-1}(\delta _1-\epsilon _1,\delta _2+\epsilon _2)\times M\times B_0$ for some small $\epsilon _2$ where there are no critical points. Let $T^1_{\partial \sigma }$ be the flow-out of the image of $\sigma \big |_{\partial ^2\mathcal {A}_1}$ . Both are rectifiable currents with the obvious orientation. The following identity of currents holds on ${f}^{-1}(\delta _1-\epsilon _1,\delta _2+\epsilon _2)\times M\times B_0$ :

$$ \begin{align*} dT^1_{\sigma}=T_{\partial \sigma}^1=(U_{{\textbf{p}}_1}\cap {f}^{-1}(\delta_1-\epsilon_1,\delta_2+\epsilon_2))\times s(s^{-1}(S_{{\textbf{p}}_1})\times B_0) \end{align*} $$

proving thus the theorem on ${f}^{-1}(\delta _1-\epsilon _1,\delta _2+\epsilon _2)\times M\times B_0$ .

From this point on we repeatedly apply Proposition 4.9 and Corollary 4.18. We notice that $(\sigma _2,\sigma _3)(\sigma _1^{-1}(S_p))$ can be substituted with $(\sigma _2,\sigma _3)(\sigma _1^{-1}(S_p)\cap N^0)$ and these points are easy to describe as $s(s^{-1}(S_p\times B_0))$ where p is any of the critical points at ith step. It is not anymore difficult to see that $T_{\sigma \big |_{\partial ^1_j N}}$ is a sum:

$$ \begin{align*} \sum_{p} (U_p\cap f^{-1}(c_i-\theta_i,c_i+\theta_i))\times s(s^{-1}(S_p\times B_0)), \end{align*} $$

where the sum here runs over the critical points that have already been crossed.▪

Proof If $H:C\times B\rightarrow M$ is a smooth map, C is an oriented, compact manifold with corners of dimension c and $\omega $ is a smooth form on M of degree $k\geq c-1$ then:

(5.1) $$ \begin{align} \int_{C}H^*d\omega+(-1)^{c-1}d\int_{C}H^*\omega=\int_{\partial C}H^*\omega. \end{align} $$

To see (5.1), apply first Stokes on $C\times B$ to

$$ \begin{align*}d(H^*\omega\wedge \pi_2^*\eta)=dH^*\omega\wedge \pi_2^*\eta+(-1)^{k}H^*\omega\wedge\pi_2^*d\eta,\end{align*} $$

where $\eta \in \Omega ^{n-k+c-1}(B)$ is a smooth test form and $\pi _2:C\times B\rightarrow B$ is the projection and then integrate over the fiber. On the closed, oriented B one has:

(5.2) $$ \begin{align}0=d\left(\int_CH^*\omega\right)\wedge \eta+ (-1)^{k-c}\left(\int_CH^*\omega\right)\wedge d\eta. \end{align} $$

Hence, using the orientation of the fiber first convention we get from (5.2)

$$ \begin{align*}(-1)^{k}\int_CH^*\omega\wedge\pi_2^*d\eta=(-1)^{k}\left(\int_CH^*\omega\right)\wedge d\eta=(-1)^{c-1}d\left(\int_CH^*\omega\right)\wedge \eta. \end{align*} $$

Take $C=\Delta ^2$ be the standard simplex in ${\mathbb R}^2$ with coordinates $(s,t)$ and on $\pi _2^*E\rightarrow C\times B$ consider the connection that “interpolates” between $\nabla _1$ , $\nabla _2$ , $\nabla _3$ :